|z-1|=|2cosθ-1+2isinθ|
=√((2cosθ-1)^2+(2sinθ)^2)
=√(4cosθ^2+1-4cosθ+4sinθ^2)
=√(5+1-4cosθ)
cosθ为-1时|z-1|最大为√10
z^2=(2cosθ+2isinθ)(2cosθ+2isinθ)
=4cosθ^2+8icosθsinθ-4sinθ^2
=4(cosθ^2-sinθ^2)-(8cosθsinθ)i
=-4cos2θ-4sin2θ
由题意知
-4sin2θ=2√3
-4cos2θ=-2
sin2θ=-(√3)/2
cos2θ=1/2
(2相限)
2θ=π/2+π/3 =2π/3
θ=π/3